AP Calculus BC 2.2 Defining the Derivative of a Function and Using Derivative Notation Study Notes - New Syllabus
AP Calculus BC 2.2 Defining the Derivative of a Function and Using Derivative Notation Study Notes- New syllabus
AP Calculus BC 2.2 Defining the Derivative of a Function and Using Derivative Notation Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Represent the derivative of a function as the limit of a difference quotient.
Key Concepts:
- Defining the Derivative of a Function and Using Derivative Notation
- Derivative as the Slope of the Tangent Line
Defining the Derivative of a Function and Using Derivative Notation
Defining the Derivative of a Function and Using Derivative Notation
The derivative of a function at a point measures the instantaneous rate of change of the function at that point. It is also the slope of the tangent line to the function’s graph at that point.
The derivative of a function \( f(x) \) at a point \( x = a \) is defined as:
\( f'(a) = \lim_{h \to 0} \dfrac{f(a + h) – f(a)}{h} \)
This limit, if it exists, gives the slope of the tangent line to the curve \( y = f(x) \) at the point \( x = a \).
General Derivative Function:
The derivative function \( f'(x) \) is defined as:
\( f'(x) = \lim_{h \to 0} \dfrac{f(x + h) – f(x)}{h} \)
This expression represents a function that gives the slope of the curve at any point \( x \) where the derivative exists.
Interpretation: The derivative tells us how fast the function is changing. A positive derivative means the function is increasing, while a negative derivative means it is decreasing.
Common Derivative Notation:
| Notation | Description |
|---|---|
| \( f'(x) \) | Leibniz’s prime notation |
| \( \dfrac{dy}{dx} \) | Derivative of \( y \) with respect to \( x \) |
| \( D_x[f(x)] \) | Differentiation operator acting on \( f(x) \) |
| \( \dfrac{d}{dx}f(x) \) | Same as above, more expanded form |
Example :
Find the derivative of the function \( f(x) = x^2 + 3x \) using the definition of the derivative.
▶️Answer/Explanation
\( f'(x) = \lim_{h \to 0} \dfrac{f(x + h) – f(x)}{h} \)
\( f(x + h) = (x + h)^2 + 3(x + h) = x^2 + 2xh + h^2 + 3x + 3h \)
\( f(x + h) – f(x) = [x^2 + 2xh + h^2 + 3x + 3h] – [x^2 + 3x] = 2xh + h^2 + 3h \)
\( \dfrac{f(x + h) – f(x)}{h} = \dfrac{2xh + h^2 + 3h}{h} = 2x + h + 3 \)
\( f'(x) = \lim_{h \to 0} (2x + h + 3) = 2x + 3 \)
Derivative as the Slope of the Tangent Line
Derivative as the Slope of the Tangent Line
The derivative of a function at a specific point represents the slope of the tangent line to the graph of the function at that point.
If a function is differentiable at a point \( x = a \), then the derivative \( f'(a) \) gives the instantaneous rate of change of the function at \( a \), and this is the same as the slope of the straight line that just “touches” the curve at \( x = a \) without cutting through it — i.e., the tangent line.
The formula is:
\( f'(a) = \lim_{h \to 0} \dfrac{f(a + h) – f(a)}{h} \)
Interpretation: – The value of the derivative tells you how steep the function is at a point. – A positive value means the function is increasing at that point. – A negative value means the function is decreasing. – A zero derivative means the tangent is horizontal at that point.
Example:
Find the equation of the line tangent to the curve \( f(x) = x^3 – 4x \) at the point where \( x = 2 \).
▶️Answer/Explanation
\( f'(x) = \dfrac{d}{dx}(x^3 – 4x) = 3x^2 – 4 \)
\( f'(2) = 3(2)^2 – 4 = 12 – 4 = 8 \)
The slope of the tangent line at \( x = 2 \) is 8.
The coordinates of the point on the curve
\( f(2) = (2)^3 – 4(2) = 8 – 8 = 0 \)
The point is \( (2, 0) \).
Use point-slope form of the line
\( y – y_1 = m(x – x_1) \)
\( y – 0 = 8(x – 2) \Rightarrow y = 8x – 16 \)
Example :
The function is \( f(x) = \sqrt{x + 1} \). Find the slope of the tangent line to the curve at the point where \( x = 3 \).
▶️Answer/Explanation
We use the chain rule to differentiate \( f(x) = \sqrt{x + 1} = (x + 1)^{1/2} \):
\( f'(x) = \dfrac{1}{2}(x + 1)^{-1/2} \cdot 1 = \dfrac{1}{2\sqrt{x + 1}} \)
\( f'(3) = \dfrac{1}{2\sqrt{3 + 1}} = \dfrac{1}{2\sqrt{4}} = \dfrac{1}{2 \cdot 2} = \dfrac{1}{4} \)
Interpretation: The slope of the tangent line to the curve at \( x = 3 \) is \( \boxed{\dfrac{1}{4}} \).